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LS 30B Mathematics for Life Scientists Homework 4 Solutions

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Choose an initial value for X and use values of r that are less than, greater than, and equal to one to test the above statements by computing two values of X for each value of r. Solution: Choose X0 = 1, r = 1: X1 = ...

Choose an initial value for X and use values of r that are less than, greater than, and equal to one to test the above statements by computing two values of X for each value of r. Solution: Choose X0 = 1, r = 1: X1 = 1 ∗ 1 = 1; X2 = 1 ∗ X1 = 1 ∗ 1 = 1: r = 2: X1 = 2 ∗ 1 = 2; X2 = 2 ∗ 2 = 4: r = 1=2: X1 = 1 2 ∗ 1 = 1 2 ; X2 = 1 2 1 2 = 1 4 : 1LS 30B Homework 4 Solutions Feb 9, 2018 5.1.3: An inedible alga is growing on a pond in a city park. Only a small part of the pond is now covered by the algae, but the area covered is doubling each day. The city decides to remove the algae once it covers half the pond. If the pond will be completely overgrown in thirty days, on what day will it be half covered? (Hint: Try working backward.) Solution: The pond will be completely overgrown in thirty days. Since the algae doubles each day the pond will be half overgrown with algae on the twenty ninth day. On the twenty eighth day it will be less than half covered. So the city will remove the algae on the twenty ninth day. 5.1.4: A rabbit population is growing at 10% a year. If there are 10 rabbits this year and time is discrete, how many will there be in 10 years? Use a loop in SageMath to check your answer. Solution: In this case r = 1:1 and X0 = 10. So X10 = (1:1)10 ∗ 10 = 25:937424601: (Round one way or the other to get the actual rabbit population.) Sage code: init = 10 r = 1.1 for i in range(0,10): x = r*init init = x init 5.1.5: While we have been working with r > 1, representing growth, r can be less than 1, representing a quantity that decreases over time. The half-life of a radioactive element is the amount of time needed for half the element to decay. What fraction of the initial amount of such an element will remain after ten half-lives? Solution: Each half life we lose 1 2 of the element. So after ten iterations we get 2110 = 1 1024. 5.1.6: When money in a bank account accrues compound interest, the interest earned in one time period is added to the principal, and then the sum is used as the base for the next time period. (a) If you start off with $1000 and earn 2% interest that is compounded annually, how much money will you have in 5 years? In 10 years? In 20 years? (b) How long will it take you to accumulate $10,000

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